PHYS852 Quantum Mechanics II, Spring 2009
HOMEWORK ASSIGNMENT 7:
1.
Spontaneous emission:
In this problem, we will use the Fermi Golden Rule to estimate the
spontaneous emission rate of an atom. We consider an atom which is initially excited, and therefore
has energy
E
e
. It decays to the ground state, having
E
g
= 0, by emitting a photon of energy
planckover2pi1
ω
.
To use the FGR,
Γ =
2
π
planckover2pi1

G
(
E
i
)

2
n
(
E
i
)
,
we need the density of photon states,
n
(
planckover2pi1
ω
), as well as the atomfield coupling constant,
G
(
planckover2pi1
ω
).
Clearly the decaying atom cannot tell whether it lives in an infinite space, or in a very large box.
Hence the decay rate should be the same in either situation, and should be independent of
L
, the
box length.
We know from mathematics that any function defined on the interval [0
,L
] which vanishes at the
edge points, can be expanded in terms of sin(
k
n
x
) where
k
n
=
nπ/L
, with
n
being any positive
integer. If the box is large enough, then the photon electric field must vanish at the walls. This field
can be therefore be expanded onto the set of ‘modes’,
u
n
x
,n
y
,n
z
(
x,y,z
) = sin(
n
x
πx/L
) sin(
n
y
πy/L
) sin(
n
z
πz/L
)
,
where the energy of each photon mode is
E
=
planckover2pi1
ck
=
planckover2pi1
c
(
π
L
)
radicalBig
n
2
x
+
n
2
y
+
n
2
z
.
a.) To determine the density of states at energy
E
, first determine
N
(
E
), which is clearly the number
of (
n
x
,n
y
,n
z
) points with energies below
E
. The density of states can then be computed via
n
(
E
) =
dN
(
E
)
dE
.
Answer:
From the box eigenmodes, we see that
n
x
,
n
y
, and
n
z
are positive integers.
Since
k
n
=
nπ/L
, we see that the spacing,Δ
k
, is given by Δ
k
=
π/L
.
Thus with
vector
k
n
x
n
y
n
z
=
π
L
(
n
x
vector
e
x
+
n
y
vector
e
y
+
n
z
vectore
z
) we see that the allowed
k
vectors lie on a 3d lattice,
with spacing
π/L
.
The question is then how many lattice points are inside a sphere of radius
k
=
E
planckover2pi1
c
, subject to
the constraint
{
k
x
,k
y
,k
z
}
>
0?
So the density of points is
ρ
=
1 point
(
π/L
)
3
=
L
3
π
3
The volume of the energy sphere is
V
(
E
) =
4
3
πk
3
(
E
)
so the number of points satisfying the two conditions is
N
(
E
) =
1
8
ρV
(
E
) =
1
8
4
L
3
k
3
(
E
)
3
π
2
=
L
3
E
3
6
π
2
planckover2pi1
3
c
3
1